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# OCR A2 CHEMISTRY F324 and F325- 14th and 22nd June 2016- OFFICIAL THREAD watch

1. (Original post by zirak46)
Hello peeps, could someone please explain how in electrochemical cells we decide which half cell is negative electrode and which is positive?

So when drawing a diagram of two half cells, which one goes on left or right?

The question is from Jan 12

Generally, The one being oxidised is negative. Vice versa.
2. I know many of you have asked this question before, but could someone elaborate on how to work this out ? [part (ii)]
Attachment 553504553506

I under standard that the Ag+ + Cl- --> agcl Calculating 2.2868/(107.9+35.5) I'd get 0.02 which is the mole for agcl. The mole ratio of cl- to agcl is 0.01 : 0.02(1:2). But how could there be 2Agcl and 1cl- ? It just doesn't make much sense to me as the equation doesn't balance up.
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3. Only acidic buffers according to spec
4. (Original post by suibster)
I know many of you have asked this question before, but could someone elaborate on how to work this out ? [part (ii)]
Attachment 553504553506

I under standard that the Ag+ + Cl- --> agcl Calculating 2.2868/(107.9+35.5) I'd get 0.02 which is the mole for agcl. The mole ratio of cl- to agcl is 0.01 : 0.02(1:2). But how could there be 2Agcl and 1cl- ? It just doesn't make much sense to me as the equation doesn't balance up.
5. (Original post by suibster)
I know many of you have asked this question before, but could someone elaborate on how to work this out ? [part (ii)]
Attachment 553504553506

I under standard that the Ag+ + Cl- --> agcl Calculating 2.2868/(107.9+35.5) I'd get 0.02 which is the mole for agcl. The mole ratio of cl- to agcl is 0.01 : 0.02(1:2). But how could there be 2Agcl and 1cl- ? It just doesn't make much sense to me as the equation doesn't balance up.
You need to assume that silver nitrate is in excess. Therefore the ratio indicates that there must have been 2 chloride ions in the original complex
6. (Original post by zirak46)
yes
You need to assume that silver nitrate is in excess. Therefore the ratio indicates that there must have been 2 chloride ions in the original complex
Assuming Ag is in excess, are you suggesting: 2Ag+ + CL- ---> 2AgCl ?
8. (Original post by zirak46)
Cu2+ + 2e- <-> Cu (s)
That is the eq on the table.
Adding H2O, decreases conc of Cu2+.
As its eq, the eq will shift to left to compensate. Therefore its more:

Cu (s) -> Cu2+ + 2e-

Hence, more oxidiation.

But the thing to understand I think is that, less Cu2+,so eq shifts to the left.

This means, the EP value decreases and so overall cell potential between Ag and Cu will increase due to the larger difference in EP between the two.

Hope that makes sense. (Im literally explaining this from wha I figured from nervous' answer, literally just understood all this from him, thought id give him a break though )
Haha. Thanks for that. And you are right by the way.

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9. (Original post by suibster)
I know many of you have asked this question before, but could someone elaborate on how to work this out ? [part (ii)]
Attachment 553504553506

I under standard that the Ag+ + Cl- --> agcl Calculating 2.2868/(107.9+35.5) I'd get 0.02 which is the mole for agcl. The mole ratio of cl- to agcl is 0.01 : 0.02(1:2). But how could there be 2Agcl and 1cl- ? It just doesn't make much sense to me as the equation doesn't balance up.
I think you may want to read the questions again. It says it uses 0.01 moles of a COMPLEX. Not of cl- ions. And this forms 0.02 moles of AgCl. This means that the complex must contain 2 cl- ions.
An example is thinking of crisps. If each crisp packet contained 3 prizes. (Wierd example lol). There would still be 1.0 mole of crisps. But if I wanted to work out moles of prizes, I would do 1.0 times 3 as there are 3 in each.
Wow I hope that makes some sense.

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10. (Original post by suibster)
I know many of you have asked this question before, but could someone elaborate on how to work this out ? [part (ii)]
Attachment 553504553506

I under standard that the Ag+ + Cl- --> agcl Calculating 2.2868/(107.9+35.5) I'd get 0.02 which is the mole for agcl. The mole ratio of cl- to agcl is 0.01 : 0.02(1:2). But how could there be 2Agcl and 1cl- ? It just doesn't make much sense to me as the equation doesn't balance up.
Think youve misunderstood pal.

It says, as did I when I first did it.

It says 0.01 mole reacts with excess Ag (however much)

First use n = mass/ mr to find mr of the precipitate

2.868/0.01 gives mr of 286.8

Now the precipitate would be AgCl (as learned from f321).

Using rhe formulae of the complexes, its either 3Cl- reacting, 2Cl- or Cl-.

As Ag is in EXCESS, all Chloride ions will react, therefore meaning either :
3AgCl, 2AgCl or AgCl is formed.

Now find Mr of each and match with Mr of 286.6 to get your answer.

As two chloride ions used, it means complex B was used
11. (Original post by ImNervous)
Haha. Thanks for that. And you are right by the way.

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Cheers pal
12. (Original post by ImNervous)
I think you may want to read the questions again. It says it uses 0.01 moles of a COMPLEX. Not of cl- ions. And this forms 0.02 moles of AgCl. This means that the complex must contain 2 cl- ions.
An example is thinking of crisps. If each crisp packet contained 3 prizes. (Wierd example lol). There would still be 1.0 mole of crisps. But if I wanted to work out moles of prizes, I would do 1.0 times 3 as there are 3 in each.
Wow I hope that makes some sense.

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That does make sense! Thank you.
13. (Original post by zirak46)
Cu2+ + 2e- <-> Cu (s)
That is the eq on the table.
Adding H2O, decreases conc of Cu2+.
As its eq, the eq will shift to left to compensate. Therefore its more:

Cu (s) -> Cu2+ + 2e-

Hence, more oxidiation.

But the thing to understand I think is that, less Cu2+,so eq shifts to the left.

This means, the EP value decreases and so overall cell potential between Ag and Cu will increase due to the larger difference in EP between the two.

Hope that makes sense. (Im literally explaining this from wha I figured from nervous' answer, literally just understood all this from him, thought id give him a break though )
Okay, suppose it just asked for the emf of Copper. By this logic, if the equilibrium shifted left then the overall cell emf becomes positive? because it's losing electrons?

Edit: Or is it because it moves to the left, the cu2+ is gaining more electrons and so the emf of the cell is negative?
14. (Original post by ImNervous)
I think you may want to read the questions again. It says it uses 0.01 moles of a COMPLEX. Not of cl- ions. And this forms 0.02 moles of AgCl. This means that the complex must contain 2 cl- ions.
An example is thinking of crisps. If each crisp packet contained 3 prizes. (Wierd example lol). There would still be 1.0 mole of crisps. But if I wanted to work out moles of prizes, I would do 1.0 times 3 as there are 3 in each.
Wow I hope that makes some sense.

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I've had a think about the question again and I believe your example is wrong, as you can not simply just change the moles of a reactant.
My conclusion is that since there's two CL- in complex B and there's 0.01 moles of B.
This means if complex b were to dissociate, it would be this:
[Co(nh3)5cl]2+2CL- ---> [Conh3)5cl]2+ + 2Cl-
So the mole ratio is 1:2 so 0.01 : 0.02

When this cl- reacts with ag+, the equation would be ag+ + cl- ---> agcl
mole ratio being 1:1 or (0.02:0.02)
15. HOW can i revise for this exam?? other than past papers and legacy papers? WANT an A! tipss pleasse
16. Can someone please tell me why the answer is -2.31. I don't understand why it's negative.
17. (Original post by KB_97)
Can someone please tell me why the answer is -2.31. I don't understand why it's negative.
Think about the standard cell potential formula

Standard cell potential = positive terminal - negative terminal

Rule: Oxidation= Negative terminal , Reduction=Positive terminal

Identify what's being oxidised and reduced.

O2 is being reduced so ------> Positive terminal

2.71 = 0.4 - (Negative terminal)

Rearrange to get Negative terminal = 0.4 - 2.71

Negative terminal (Al) = -2.31 V
18. (Original post by KB_97)
Can someone please tell me why the answer is -2.31. I don't understand why it's negative.
Whichever reaction is being reduced is the largest electrode potential value.

Cell potential is Largest Ep - Smallest Ep value

Hence: 0.4 - (Ep of other half cell) = 2.71
Rearrange to get Ep of ther half cell = -2.71
19. (Original post by tcameron)
whenever we do kc calculations the inital moles of the reactants is always known as that's what is added to make the products so the initial moles of the products is always 0
so that's why you needed to use the products as their initial moles were 0
all it was was 0.168x3
i'm confused again

we already know the moles for the reactant is three?
20. (Original post by suibster)
I've had a think about the question again and I believe your example is wrong, as you can not simply just change the moles of a reactant.
My conclusion is that since there's two CL- in complex B and there's 0.01 moles of B.
This means if complex b were to dissociate, it would be this:
[Co(nh3)5cl]2+2CL- ---> [Conh3)5cl]2+ + 2Cl-
So the mole ratio is 1:2 so 0.01 : 0.02

When this cl- reacts with ag+, the equation would be ag+ + cl- ---> agcl
mole ratio being 1:1 or (0.02:0.02)
That's right. That's what I was trying to achieve with my *****y example lol.

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